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Calculus, 2ADV C2 2022 HSC 25

Let  `f(x)=sin(2x)`.

Find the value of `x`, for  `0 < x < pi`, for which  `f^(′)(x)=-sqrt3`  AND  `f^(″)(x)=2`.  (3 marks)

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`(7pi)/12`

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`f^(′)(x)=2cos(2x)`

`2cos(2x)` `=-sqrt3`  
`cos(2x)` `=- sqrt3/2`  
`2x` `=pi-pi/6,\ \ pi+pi/6`  
  `=(5pi)/6,\ \ (7pi)/6`  
`x` `=(5pi)/12,\ \ (7pi)/12`  

 
`f^(″)(x)=-4sin(2x)`

`-4sin(2x)` `=2`  
`sin(2x)` `=- 1/2`  
`2x` `=pi+pi/6,\ \ 2pi-pi/6`  
  `=(7pi)/6,\ \ (11pi)/6`  
`x` `=(7pi)/12,\ \ (22pi)/12`  

 
`:.x=(7pi)/12\ \ text{(satisfies both equations)}`


Mean mark 53%.

Calculus, 2ADV C2 2006 HSC 2aii

Differentiate  `sin x/(x + 1)`  with respect to `x`.  (2 marks)

 

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`dy/dx = {cos x (x + 1) – sin x} / (x + 1)^2`

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`y = sin x / (x + 1)`

`text(Using)\ \ d/dx (u/v) = (u^{\prime} v – uv^{\prime})/v^2`

`u` `= sin x` `v` `= x + 1`
`u^{\prime}` `= cos x` `\ \ \ v^{\prime}` `= 1`

 

`:.dy/dx = {cos x (x + 1) – sin x} / (x + 1)^2`

Calculus, 2ADV C2 EQ-Bank 10

The function  `f(theta) = sin^3(2 theta)`.

If  `f′(theta) = 6 cos(2 theta) - 6 cos^n (2 theta)`, find the value of `n`.  (2 marks)

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`3`

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`f(theta)` `= sin^3(2 theta)`
  `= (sin(2theta))^3`

 

`f′(theta)` `= 3 xx 2cos(2 theta) xx sin^2(2 theta)`
  `= 6 cos(2 theta)(1 – cos^2(2 theta))`
  `= 6 cos (2 theta) – 6 cos^3(2 theta)`

 
`:. \ n = 3`

Calculus, 2ADV C4 2018 HSC 5 MC

What is the derivative of  `sin(ln x),` where  `x > 0`?

  1. `cos (1/x)`
  2. `cos (ln x)`
  3. `cos ((ln x)/x)`
  4. `(cos (ln x))/x`
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`D`

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`y` `= sin (ln x)`
`(dy)/(dx)` `= cos (ln x) xx d/(dx) (ln x)`
  `= cos (ln x) xx 1/x`
  `= (cos (ln x))/x`

 `=>  D`

Calculus, 2ADV C2 2011 HSC 4a

Differentiate  `x/sinx`  with respect to  `x`.   (2 marks)

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 `(sin x\ – x cos x)/(sin^2x)`

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`y = x/sinx`

`u = x` `\ \ \ \ \ u prime = 1`
`v = sin x` `\ \ \ \ \ v prime = cos x`

 

`text(Using)\ \ d/dx (uv) = (u prime v\ – uv prime)/(v^2),`

`dy/dx` `= (1 * sinx \ – x * cos x)/((sin x)^2)`
  `= (sin x\ – x cos x)/(sin^2x)`